I think it's more elegant to do the indefinite integral in polar coordinates, recognizing this is the upper half of the circle with radius=2 centered at the origin as in

integral sqrt(4-x^2) = integral integral 2r dr dT

where r = radius and T = theta = angle formed with the positive x axis. The conversion to polar coordinates uses r^2 = (x/2)^2 + (y/2)^2, and we know from the original eqn that r=2. The extra r in the double integral is from the Jacobian necessary in converting to polar coord. Then finding this integral is easy:

= integral r^2 dT = Tr^2 = 4T + C

which I guess would work for finding areas of sectors of the circle of radius 2.

What I'm curious is if I convert back to x and y, if I get the solution above, using x = rcosT, and r^2 = (x/2)^2 + (y/2)^2

I think it's more elegant to do the indefinite integral in polar coordinates, recognizing this is the upper half of the circle with radius=2 centered at the origin as in

integral sqrt(4-x^2) = integral integral 2r dr dT

where r = radius and T = theta = angle formed with the positive x axis. The conversion to polar coordinates uses r^2 = (x/2)^2 + (y/2)^2, and we know from the original eqn that r=2. The extra r in the double integral is from the Jacobian necessary in converting to polar coord. Then finding this integral is easy:

= integral r^2 dT = Tr^2 = 4T + C

which I guess would work for finding areas of sectors of the circle of radius 2.

What I'm curious is if I convert back to x and y, if I get the solution above, using x = rcosT, and r^2 = (x/2)^2 + (y/2)^2

It doesn't :-(
You are trying to calculate
integral sqrt(4-x^2) in the same way you compute the definite integral
int (-infty +infty) exp(-x^2)!

But here the result we are looking for is the antiderivatives of sqrt(4-x^2)

You are mixing (and confusing) two concepts: area and antiderivative.
It is true that you can calculate area using the fundamental theorem of calculus which states that the area under a curve f(x) in the interval [a,b] is F(b) - F(a), where F'(x) = f(x). But here F(x) is the primitive function we are looking for.
A function such that F'(x)=sqrt(4-x^2)

(2·arcsin(x/2) + x·sqrt(4 - x^2)/2)' = sqrt(4-x^2)

April 22nd 2005, 02:00 PM

billh

OK, I can see that. Thanks for the clarification. But, if the problem HAD been a definite integral from -2<=x<=2, I COULD have done it with polar coordinates! We just finished doing that in class, so I was "sensitized" to looking for equations of circles.

April 22nd 2005, 04:34 PM

ticbol

billh,

I see your point. You are correct in a way.

[sqrt(4 -x^2)]dx can be viewed as the dA for the area above the x-axis of the circle centered at the origin with radius = 2 units.

Now, using polar coordinates,
The said whole circle centered at origin with radius 2 is
r = 2 ---equation of the whole circle.

If we need to find the area of the said circle above the "equivalent of x-axis" or the area from (theta = 0) to (theta = pi), then we get the dA first.

dA is an infinitesimal sector of the circle, whose radius is 2 and whose central angle is dtheta or dT.
So its subtended arc is (radius)(central angle) = 2*dT = 2dT
dA then is (1/2)(radius)(subtended arc) = (1/2)(2)(2dT) = 2dT

-------------
But then, the original question is for integrating sqrt(4 -x^2) dx,
where sqrt(4 -x^2) can be any quantity in general---not necessarily the positive y-coordinate of a circle as mentioned above.

April 28th 2005, 07:35 AM

billh

I mentioned this problem to my calculus professor (2nd year) and he looked at me like I was an idiot. I described the problem as an "indefinite integral", which is the term I learned in Calc I. He said "don't say 'indefinite integral' just think function. It is just the antiderivative and has nothing to do with finding areas", in and of itself, ie apart from the the other part of the Fund Thm of Calc. I looked like and idiot, but at least I learned something.